8.821 F2008 Lecture 14: Wave equation in AdS, Green’s function
Lecturer: McGreevy Scribe: Mark Hertzberg
November 11, 2008
Topics for this lecture
• Find φ[φ0 ] (x) by Green functions in x-space (efficient)
• Compute hOOi, counter terms
• Redo in p-space (general)
References
• Witten, hep-th/9802150
• GKP, hep-th/9802109
Solving Wave Equation I (Witten’s method)
Let’s study the wave equation in AdS in some detail. This first method uses a trick by Witten
which is efficient but slightly obscure.
If we know “bulk-to-boundary” Green’s function K regular in the bulk, such that
(− + m2 )Kp (z, x) = 0
Kp (z, x) →
∆− δD (x
− p), z → where p is some point on the boundary,
then the field in the bulk
Z
[φ0 ]
0
∆− Ren
φ (z, x) = dD x0 φRen
φ0 (x)
0 (x )Kx0 (z, x) → z
solves (1).
1
(1)
(2)
Euclidean AdS
Recall the metric on AdS with curvature scale L in the upper half plane coordinates:
ds2 = L2
dz 2 + dx2
z2
Now here comes some fancy tricks, thanks to Ed:
Trick (1): Pick p = “point at ∞”. This implies that the Green’s function K∞ (z, x) is x-independent.
The wave equation at k = 0:
0 = −z D+1 ∂z z −D+1 ∂t + m2 K∞ (z)
can easily be solved. The solution is power law (recall that in the general-k wave equation, it was
the terms proportional to k 2 that ruined the power-law behavior away from the boundary)
K∞ (z) = c+ z ∆+ + c− z ∆−
We can eliminate one of the constants: c− = 0, whose justification will come with the result.
Trick (2): Use AdS isometries to map p = ∞ to finite x. Let xA = (xµ , z), take xA → (x0 )A =
xA /(xB xB ). The inversion of this mapping is:
(
µ
xµ → z 2x+x2
I:
z
z → z 2 +x
2
Claim: I
A) is an isometry of AdS (also Minkowski version, see pset 4)
B) is not connected to 1 in SO(D,2)
C) maps p = ∞ to x = 0, i.e., I : K∞ (z, x) → K∞ (z 0 , x0 ) = K0 (z, x) = c+ z ∆+ /(z 2 + x2 )∆+ .
Some notes:
(i) That this solves the wave equation (1) as neccessary can be checked directly.
(ii) The Green’s function is
Kx0 (z, x) = c+
z ∆+
≡ K(z, x; x0 )
(z 2 + (x − x0 )2 )∆+
(iii) The limit of the Green’s function as z → 0 , i.e. the boundary is
(
cz ∆+ → 0, if x 6= x0
0
K(z, x; x ) →
cz −∆+ → ∞, if x = x0
2
(recall that ∆+ > 0 for any D, m). More specifically, the Green’s function approaches a delta
function:
K(z, x; x0 ) → const · ∆− δ D (x − x0 ).
Clearly it has support only near x = x0 , but to check this claim we need to show that it has finite
measure:
Z
Z
c∆+ −∆−
dD x 2
dD x −∆− K0 (, x) =
( + x2 )∆+
Z
1
cD 2∆+ −D
dD x̄
=
2∆
+
(1 + x̄2 )∆+
D
π 2 Γ(∆+ −
= c
Γ(∆+ )
D
2)
.
We will choose the constant c to set this last expression equal to one. Hence,
Z
[φ0 ]
0
φ (z, x) =
dD x0 Kx0 (z, x)φRen
0 (x )
Z
x ∆+
=
dD x0 c 2
φRen (x0 ) ;
(z + (x − x0 )2 )∆+ 0
this solves (1) and approaches ∆− φRen
0 (x) as z → .
The action is related to expectation values of operators on the boundary:
i
h
R
S φ[φ0 ] = − lnhe− φ0 O i
Z
η
√
= −
γ φ n · ∂φ
2 ∂AdS
Z
η
D √
zz
= −
d x g g φ(z, x)∂z φ(z, x)
2
z=
Z
η
Ren
dD x1 dD x2 φRen
= −
0 (x1 )φ0 (x2 )F (x1 , x2 )
2
where the “flux factor” is
Z
F (x1 , x2 ) ≡
dD x
K(z, x; x1 )z∂z K(z, x; x2 ) .
zD
z=
The boundary behavior of K is:
∆+
0 ∆−
D
0
2
∆+
K (z, x; x )
=
δ (x − x ) + O( ) + z=
c
2
+ O( )
(x − x0 )2∆−
D
the first terms sets: c−1 = π 2 Γ(∆+ − D
2 )/Γ(∆+ ), the second term is subleading in z.
1
z∂z K(z, x; x0 )
= ∆+ ∆− δ(x − x0 ) + ∆+ cz ∆+
+ ...
(x − x0 )2∆+
z=
Ok, now for the 2-point correlation function on the boundary:
h
i
δ
δ
G2 (x1 , x2 ) ≡ hO(x1 )O(x2 )ic =
−S φ[φ0 ] = ηF (x1 , x2 ).
δφ0 (x1 ) δφ0 (x2 )
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We must be careful when evaluating the cases x1 6= x2 and x1 = x2 , which we do in turn.
Firstly, if x1 6= x2 :
G2 (x1 6= x2 ) =
=
=
Z
∆+ cz ∆+
η
2
D
−D
∆− D
2
d xz
z δ (x − x1 ) + O(z ) (ignore by x1 6= x2 ) +
+ O(z )
2 z=
(x1 − x2 )2∆+
1
η
+ O(2 )
c∆+ −D+∆− +∆+
2
(x1 − x2 )2∆+
ηc∆+
.
2(x1 − x2 )2∆+
Good. This is the correct form for a two point function of a conformal primary of dimension ∆+
in a CFT; this is a check on the prescription.
Secondly, if x1 = x2 :
G2 (x1 , x2 ) = η ∆− 2∆− −D δ D (x1 − x2 ) +
c∆+
+ ∆+ c2 2∆+ −D
(x1 − x2 )2∆+
Z
1
d x
2∆
+
(x − x1 )
(x − x2 )2∆+
D
As → 0, the first term is divergent, the second term is finite, and the third term vanishes. The
first term is called a “divergent contact term”. It is scheme-dependent and useless.
Remedy: Holographic Renormalization. Add to Sgeometry the contact term
Z
η
2
dD x −∆− 2∆− −D (φRen
(x))
∆S = Sc.t. =
0
2 bdy
Z
η
√ 2
= −∆−
γ φ (z, x).
2 ∂AdS,z=
Note that this doesn’t affect the equations of motion. Nor does it affect G2 (x1 6= x2 ).
Solving Wave Equation II (k-space)
Since the previous approach isn’t always available (for example if there is a black hole in the
spacetime), let’s redo the calculation in k-space.
Return to wave equation
0 = z D+1 ∂z z −D+1 ∂z − m2 L2 − z 2 k 2 fk (z)
with k 2 = −ω 2 + k2 > 0. The solution is
D
D
fk (z) = AK z 2 Kν (kz) + AI z 2 Iν (kz),
p
with ν = (D/2)2 + m2 L2 = ∆+ − D/2. Assume k ∈ R (real time issues later). As z → ∞:
Kν ∼ e−kz and Iν ∼ ekz . The latter is not okay, so AI = 0.
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At boundary:
(
Kν (n) ∼ n
−ν
2
4
a0 + a1 n + a2 n + . . . +
nν (b0 + b1 n2 + b2 n4 + . . .), ν ∈
/R
nν ln n (b0 + b1 n2 + b2 n4 + . . .), ν ∈ R
Hence
D
fk (z) = AK z D/2 Kν (kz) ∼ z 2 ±ν = z ∆± , as z → 0.
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